# Proof: A tangent space of the manifold of SPD matrices is the set of symmetric matrices

The set of SPD matrices, $$\mathbb{P}_n := \{X \in \mathbb{R}^{n \times n} | X=X^T, X \succ 0 \}$$, forms a differentiable manifold.

Claim: The tangent space at a point, $$A, T_A\mathcal{P}_n$$ is the space of symmetric matrices.

Info: I have seen this mentioned in a paper (https://www.ncbi.nlm.nih.gov/pubmed/27845666), but (with my limited understanding of differential geometry) can't see how this occurs (although it may actualy be obvious to someone with a more rigorous maths education). I have also checked out the book they referenced when making this claim (http://www.cmat.edu.uy/~lessa/tesis/Positive%20Definite%20Matrices.pdf) in order to understand it. Sadly the section that I think they are referencing (Start of chapter 6) is far too formal for me. For completeness I will write out what is written there, and if it is the proof I am looking for then an explanation of some of the jumps in the steps would answer my question.

1) The space $$\mathbb{M}_n$$ is a Hilbert space with inner product $$\langle A,B \rangle = tr A^*B$$ and the associated norm $$\vert \vert A \vert \vert _2 = (tr A^*A)^{\frac{1}{2}}$$

2) The set of Hermitian matrices constitutes a real vector space $$\mathbb{H}_n$$ in $$\mathbb{M}_n$$.

3) The subset $$\mathbb{P}_n$$ consisting of strictly positive matrices is an open subset in $$\mathbb{H}_n$$. Hence it is a differentiable manifold.

4) The tangent space to $$\mathbb{P}_n$$ at any of its points A is the space $$T_A\mathbb{P}_n = \{A\} \times \mathbb{H}_n$$

If this is indeed the proof that I seek then these are the steps that I don't understand:

3) I know that open subsets of manifolds are manifolds, and I am also presuming that "stricly positive" means positive definite. But how do I know that this manifold is differentiable, ie has differentiable transistion maps?

4) This is my main confusion - $$\mathbb{H}_n$$ is the set of symmetric matrices, but why is the tangent space at any point defined like this? I understand the tangent space to be the collection of velocity vectors through that point. Why can I represent the gradients of all the geodesics that pass through a point like this?

EDIT - UPDATE:

Having been browsing the stack I also came across tangent space clarification where someone is asking why tangent spaces aren't just defined in the same way (ish) as they were in 4) - does anyone know where this definition/way of writing has come from?

4) The tangent space $$T_a V$$ to a vector space $$V$$ (in this case $$V=\mathbb{H}(n)$$) can be identified with the vector space itself (via the isomorphism which takes an element $$v\in V$$ to the directional derivative $$D_v\vert_a$$). Moreover, the tangent space to an open subset of a manifold is isomorphic to the tangent space of the manifold itself. Hence in our case we have $$T_A\mathbb{P}(n) \cong T_A \mathbb{H}(n) \cong \mathbb{H}(n)\,.$$ For reference I suggest you read the beginning of the chapter "Tangent vectors" in the book "Introduction to Smooth Manifolds" by John M. Lee.